(a) To express \(f(x) = 2x^2 - 16x + 23\) in the form \(2(x + a)^2 + b\), complete the square:

Start with \(2(x^2 - 8x) + 23\).

Complete the square for \(x^2 - 8x\):

\(x^2 - 8x = (x-4)^2 - 16\).

Thus, \(2(x^2 - 8x) = 2((x-4)^2 - 16) = 2(x-4)^2 - 32\).

So, \(f(x) = 2(x-4)^2 - 32 + 23 = 2(x-4)^2 - 9\).

(b) The vertex form \(2(x-4)^2 - 9\) shows the minimum value is \(-9\) when \(x = 4\). Since \(x < 3\), the range is \(y > -7\).

(c) To find \(f^{-1}(x)\), start from \(y = 2(x-4)^2 - 9\).

\((x-4)^2 = \frac{y+9}{2}\).

\(x-4 = \pm \sqrt{\frac{y+9}{2}}\).

Since \(x < 3\), choose the negative root: \(x = 4 - \sqrt{\frac{y+9}{2}}\).

Thus, \(f^{-1}(x) = 4 - \sqrt{\frac{x+9}{2}}\).